Respuesta :
Answer:
Explanation:
Initially the two boys were sitting on the periphery , total moment of inertia
= 1/2 M r² + 2mr² ; M is mass of the merry go round , m is mass of each boy and r is the radius
1/2 x 200 x 2² + 2 x 40 x 2²
= 400 + 320
I₁ = 720 kg m²
Finally the two boys were sitting at the middle , total moment of inertia
= 1/2 M r² + 2m( r/2)² ; M is mass of the merry go round , m is mass of each boy and r is the radius
1/2 x 200 x 2² + 2 x 40 x 1²
= 400 + 80
I₂ = 480
Now the system will obey law of coservation of angular momentum because no torque is acting on the system.
I₁ω₁ = I₂ω₂ , I₁ and ω₁ are moment of inertia and angular velocity of first case and I₂ and ω₂ are of second case.
720 X 12 = 480 ω₂
ω₂ = 18 rad / s
The final angular speed of the merry-go-round - children system is 18 rad/s.
- The given parameters;
- mass of a child, m = 4 0 kg
- mass of the merry-go-round, M = 200 kg
- radius of the merry-go-round, r = 2 m
- initial angular speed, ω₁ = 12 rad/s
Apply the principle of conservation of angular momentum as shown below;
[tex]I_1 \omega _1 = I_2 \omega _2\\\\(\frac{1}{2} Mr^2 \ + 2mr^2)\omega_1 = (\frac{1}{2} Mr^2 \ + 2m(\frac{r}{2} )^2)\omega_2 \\\\(\frac{1}{2} M \ + 2m)\omega_1 r^2 =(\frac{1}{2} M \ + \frac{1}{2} m)\omega_2 r^2\\\\(\frac{1}{2} M \ + 2m)\omega_1 =(\frac{1}{2} M \ + \frac{1}{2} m)\omega_2 \\\\(\frac{1}{2} \times 200 \ + \ 2\times 40)\times 12 = (\frac{1}{2} \times 200 \ + \frac{1}{2} \times 40)\omega_2\\\\2160 = 120\omega_2 \\\\\omega _2 = \frac{2160}{120} \\\\\omega _2 = 18 \ rad/s[/tex]
Thus, the final angular speed of the merry-go-round - children system is 18 rad/s.
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